As you intimated, the radiated heat Energy output of an object is described by the Stefan-Boltzmann Law, which is E = [Object Temp ]^4 * [Stefan-Boltzmann Constant]

However, Temp must be in units of an absolute temperature scale, typically Kelvin.

So the relative heat output of a 90C vs 20C objects will be (translating to K):

383^4 / 293^4 = 2.919x

Plugging in the constant (5.67 * 10^-8 W/(m^2*K^4)) The actual values for heat radiation energy output for objects at 90C and 20C objects is 1220 W/m^2 and 417 W/m^2

The incidence of solar flux must also be taken into account, and satellites at LEO and not in the shade will have one side bathing in 1361 W/m^2 of sunlight, which will be absorbed by the satellite with some fractional efficiency -- the article estimates 0.92 -- and that will also need to be dissipated.

The computer's waste heat needs to be shed, for reference[0] a G200 generates up to 700W, but the computer is presumably powered by the incident solar radiation hitting the satellite, so we don't need to add its energy separately, we can just model the satellite as needing to shed 1361 W/m^2 * 0.92 = 1252 W/m^2 for each square meter of its surface facing the sun.

We've already established that objects at 20C and 90C only radiate 1220 W/m^2 and 417 W/m^2, respectively, so to radiate 1252 W per square meter coming in from the sun facing side we'll need 1252/1220 = 1.026 times that area of shaded radiator maintained at a uniform 90C. If we wanted the radiator to run cooler, at 20C, we'd need 2.919x as much as at 90C, or 3.078 square meters of shaded radiator for every square meter of sun facing material.

[0] Nvidia G200 specifications: https://www.nvidia.com/en-us/data-center/h200/